Find f(h(x)) f(x)=2x^3+8 h(x) = cube root of 12-5x

$f (x) = 2 x^{3} + 8$ $h (x) = \sqrt[3]{12 - 5 x}$

Step 1

Set up the composite result function.

$f (h (x))$

Step 2

Evaluate $f (\sqrt[3]{12 - 5 x})$ by substituting in the value of $h$ into $f$ .

$f (\sqrt[3]{12 - 5 x}) = 2 {(\sqrt[3]{12 - 5 x})}^{3} + 8$

Step 3

Simplify each term.

Rewrite ${\sqrt[3]{12 - 5 x}}^{3}$ as $12 - 5 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{12 - 5 x}$ as ${(12 - 5 x)}^{\frac{1}{3}}$ .

$f (\sqrt[3]{12 - 5 x}) = 2 {({(12 - 5 x)}^{\frac{1}{3}})}^{3} + 8$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\sqrt[3]{12 - 5 x}) = 2 {(12 - 5 x)}^{\frac{1}{3} \cdot 3} + 8$

Combine $\frac{1}{3}$ and $3$ .

$f (\sqrt[3]{12 - 5 x}) = 2 {(12 - 5 x)}^{\frac{3}{3}} + 8$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{12 - 5 x}) = 2 {(12 - 5 x)}^{\frac{3}{3}} + 8$

Rewrite the expression.

$f (\sqrt[3]{12 - 5 x}) = 2 (12 - 5 x) + 8$

Simplify.

$f (\sqrt[3]{12 - 5 x}) = 2 (12 - 5 x) + 8$

Apply the distributive property.

$f (\sqrt[3]{12 - 5 x}) = 2 \cdot 12 + 2 (- 5 x) + 8$

Multiply $2$ by $12$ .

$f (\sqrt[3]{12 - 5 x}) = 24 + 2 (- 5 x) + 8$

Multiply $- 5$ by $2$ .

$f (\sqrt[3]{12 - 5 x}) = 24 - 10 x + 8$

Step 4

Add $24$ and $8$ .

$f (\sqrt[3]{12 - 5 x}) = - 10 x + 32$